Question Paper Solutions
Unit 1
1. Define a) Young’s Modulus, b) Shear Modulus and c) Poisson’s Ratio and write the
relationship between them
( June 2014)
Solution:
Young’s Modulus or Modulus of Elasticity:
It states that “Within the elastic limit the stress is directly proportional to strain”
i)
i,e. Stress α strain
tress = Strain x constant
=
This constant is called Elastic modulus or Modulus of Elasticity or Young’s Modulus and it is
denoted as E
E
ii)
Shear Modulus or Modulus of Rigidity:
It is defined as "the ratio of shear stress to the displacement per unit sample length (shear
strain)" and denoted by G or C or N expressed in kN /m2.
τ=
iii) Poisson’s Ratio :
Poisson’s Ratio is defined as the ratio betw
een lateral str ain to the longitudinal
strain and denoted by µ or 1/m. It isdimensionless quantity.
µ or 1/m =
Relation between them is =
(
+)
=
2. Define Bulk Modulus and Volumetric Strain
(June 2013 /Jan2015/June2015)
Bulk Modulus: The ratio fo direct stress to volumetric strain.
K
=
=
Volumetric Strain : It is defined as theratio etw
b een change in volume to the actual volume
=
=
3. Derive an equation for deformation in tapering circular bar subjected to an axial
load P
(June 2014)
P
P
x
dx
L
Consider a tapering bar as shown ni the fig. The diamete r of the bar varies from D to d over a
length of L
The diameter at the section AB =
−
=
=
The change in length over the el ement dx =
The total change in length dl = ∫
(
)
=
−
(
…….(1 )
)
=
(
)
−
=
4. Breifly explain the behavior of ductile material under gradually increasing tensile
load
(July2014/Jan2015)
A typical tensile test specimen on mild steel is as shown above. The ends of the specimen are
gripped into a universal testing machine. The specimen has lager diameter at the ends to see
that the specimen at the ends to see that the specimen does not fail in the end regions. The
specimen should fail in the gauged portion. The deformation is recorded as the load is
applied(increases).
The elongation is recorded with the help of strain gauges. The loading is done till the
specimen fails. A graph of stress verses strain is plotted and a typical stress strain curves for
mild steel is as follows
Point A -Proportionality Limit
Point B- Elastic Limit
Point C- Lower Yield point
Point C’-Upper yield Point
Point D- Ultimate stress
Point E- Failure or Rupture or Breaking stress
From the graph the plot from O-A is linear i,e the stress is proportional to the strain within
this limit as such A is called proportionality limit. “Hooks Law” is valid in this region. On
further increase in load, the curve takes a fall as represented from A to C. Between A and C
there exists a point B. The material behaves like an elastic material till this limit and as such
B is called Elastic Limit. Loading the material beyond C causes permanent deformation of
the specimen i,e the cross sectional area reduces and length increases till the upper yield
point C is reached. Loading the specimen beyond C, the material regains some strength and
this continues till ultimate stress (D) is reached. On further increase in load the specimen
finally fails at failure stress(E).
1) Proportional Limit / Limit of proportionality :It is the limiting or maximum stress value upto which the stress is proportional to the
strain.
2) Elastic Limit :-This is the maximum or limiting stress value such that there is mo
permanent deformation in the material.
3) Yield Point /Yield Strength:- These are the lower stresses at which the extension of the
specimen is rapid without much increase in the strain.
4) Ultimate stress or yield strength:- this is the maximum stress, the material can resist.
5) Breaking stress/ Breaking strength:- this is stress at which the specimen fails or breaks or
ruptures.
5. A member is of total length of 2m its diameter is 40mm for the first 1m length. In
the next 0.5m it is gradually decreases to a diameter ‘d’. for the remaining length
the diameter ‘d’ remains same. The member is subjected to an axial pull of 150 kN
and its total elongation is 2.39mm. Determine the diameter ‘d’, assume E = 2 X
105N/mm2
(Jan2015)
Solution:
P
d=40m m
=
2.39 =
+
+
×
×
×
×
+
×
× ×
+
×
×
Solving d =20mm
6. A member ABCD subjected to point loads P1, P2, P3 and P4 as shown in fig 1.
Calculate P2 for equilibrium if P1 = 45 kN, P3 = 450kN and P4 = 130kN, if E = 2.1X
105 N/mm2. Determine the total elongation of the member.
(June 2013)
625mm2
P1
2500mm2
P3
P2
1250mm2
600mm
900mm
1200mm
P4
P1 = 45 kN, P3 = 450 kN and P4 = 130 kN
From Figure P1+ P3 = P2+ P4  P2 = 365
45 kN
45kN ( 365= 320+45)
(450 = 320+130)
320kN
320kN
130kN
=
==
45 × 10 × 1200
2.1 × 10
625
dl =
130kN
−
−
+
320 × 10 × 600
625
+
130 × 10 × 1200
900
7. The bar shown in fig 2 is tested in universal testing machine. It is observed that at a
load of 40kN the total extension is 0.285mm. Determine the Young’s modulus of the
material
(Dec 2014)
40kN
d1 = 25mm
d2 =20mm
160mm
240mm
=
0.285 =
d3 = 25mm
40 × 10
+
4 × 160
× 625
40kN
160m
+
4 × 240
× 400
4 × 160
× 625
E = 56.63X 103 N/mm2
8. A mild steel rod 2.5 m long having a cross sectional area of 50 mm2 is subjected to a
tensile force of 1.5 kN. Determine the stress, strain, and the elongation of the rod.
Take E = 2 × 105 N/mm2
(June 2015)
Solution:
Data Given
Length of the rod ‘L’ = 2.5 m = 250 mm
Area of cross-section ‘A’ = 50 mm2
Tensile force ‘P’ = 1.5 kN = 1.5 × 103 N
Young’s Modulus ‘E’ = 2 × 105 N/mm2
Stress σ = P/A= 1.5×103 / 50 =30N /mm2
Since, E = Stress / Strain Strain = Stress / E = 30 / 2 × 105 = 0.0015
Also, Elongation = Strain x Original length = 0.0015 × 2500 = 0.375 mm.
Unit 2
1. Define composite member, temperature stresses
(Dec 2013)
A section made up of more than one material designed to resist the applied load is called
'COMPOSITE SECTION'
When a body is subjected to change in temperature its dimensions will also be changed
(since the bodies are subjected to thermal expansion or contraction). For metals when the
temperature of a body is increased there is a corresponding increase in its dimensions. When the
body is allowed to expand (without restraining) no stress develops. But, in case the body is
restrained prevents the expansion, then the stresses in the body will develop. These stresses are
called as thermal stress
2. Define Modular Ratio, Volumetric strain
(Jun 2013/June2015)
Modular Ratio: Modular ratio between two materials is defined as the ratio of Young's Modulus
of Elasticity of two materials.
Volumetric Strain : It is defined as theratio etw
b een change in volume to the actual volume
=
=
3. Derive the relation between modulus of rigidity, young’s Modulus and poisson’s ratio
(Dec 2013)
Relation between E, G and
:
Let us establish a relation among the elastic constants E,G and u. Consider a cube of material of
side ‘a' subjected to the action of the shear and complementary shear stresses as shown in the
figure and producing the strained shape as shown in the figure below.
Assuming that the strains are small and the angle A C B may be taken as 450.
Therefore strain on the diagonal OA
= Change in length / original length
Since angle between OA and OB is very small hence OA @ OB therefore BC, is the change in
the length of the diagonal OA
Now this shear stress system is equivalent or can be replaced by a system of direct stresses at
450 as shown below. One set will be compressive, the other tensile, and both will be equal in
value to the applied shear strain.
thus, for the direct state of stress system which applies along the diagonals:
4. Define principle of superposition
(June 2014)
The homogeneity and additivity properties together are called the superposition principle.
A linear function is one that satisfies the properties of superposition. Which is defined as
Additivity
Homogeneity
for scalar a.
5. A steel rod is 18m long at a temperature at 25˚C. Find the free expansion when the
temperature is raised to 85˚C. Also find the temperature stress produced when
i) the expansion is completely prevented
ii) The rod is permitted to expand by 4.5mm
Given Es = 200KPa, s= 12X 10-6 /˚C
(June 2015)
Solution:
i) When the expansion is complet
ely p revented
-6
The free expansion l= L.α.Δ
t = 18 (85-25)1210 = 0.01296m
The thermal stress is = σt = E = E L t/L = E  t = 200x103x12x10-6(85-25)
2
= 144 N/mm
ii) when the rod is permitted to expand by 4.5mm
the expansion = 0.01296X1000-4.5 = 8.46mm
the thermal stresses = E (L- δ) t/L = 6.78N/mm2
6. A bar of brass of 25 mm diameter is enclosed in a steel tube of 25mm internal diameter
and 50mm external diameter. The bar & the tube are rigidly connected at both ends. Find
the stresses in both the materials of the system, where the temperature is raised from 15 ˚C
to 95˚C. Assume Es = 2X 105MPa, s= 11.6 X 10-6 /˚C, Eb= 1X 105MPa, b= 18.7 X 10-6
(Dec 2013)
/˚C
Solution: i) When the expansion iscompletely prevented
The free expansion =l L.α.Δt = 18 (85-25)1210-6 = 0.01296m
The thermal stress is = σt = E = E L t/L = E  t = 200x103x12x10-6(85-25)
= 44
1 N/mm2
ii) when the rod is permitted to expand by 4.5mm
the expansion = 0.01296X1000-4.5 = 8.46mm
the thermal stresses = E (L- δ) t/L = 6.78N/mm2
7. A load of 2MN is applied on a column 500mm X 500mm. The column is reinforced with
four steel bars 10mm diameter, on
e in each corner. Find the stresses in concrete and steel
5
2
bar. Es = 2 .1X 10 N/mm , Ec= 1.4X 104 N/mm2
Solution :
=
=
.
= 2 × 10
Strength of materials
10CV33
= 6.58 × 10
= 0.303
µ=
×
E= =
.
=
=
×
× ×
×
= 88. 42 × 10
/
E =3K( 1-2µ)
K=
(
)
= 74.80 × 10
/
E = 2G(1+µ )
G=
(
)
= 33.93 × 10
/
8. A 12mm diameter specimen is subjected to a tensile force of 20kN and deformation is
0.3mm, observed over a gauge length of 150mm. The reduction in diameter is 0.0079mm.
Determ ine th
(June2014)
e elastic constants
Solution :
=
=
=
.
= 2 × 10
.
=
= 6.58 × 10
µ = = 0.303
E ==
×
×
= 88.42 × 10
×
× ×
/
E =3
K(1-2µ)
K =
(
)
= 74.80 × 10
/
E = 2G(1+µ)
G =
(
)
= 33.93 × 10
/
Unit 3
1. Define principal stresses and principal planes.
(Jan2015)
For a given compound stress system, there exists a maximum normal stress and a
minimum normal stress which are called the Principal stresses. The planes on which these
Principal stresses act are called Principal planes. In a general 2-D stress system, there are two
Principal planes which are always mutually perpendicular to each other. Principal planes are free
from shear stresses. In other words Principal planes carry only normal stresses.
Dept of Civil Engg., SJBIT
Page 10
Strength of materials
10CV33
2. In a 2-D stress system compressive stresses of magnitudes 100 MPa and 150 MPa act in
two perpendicular directions. Shear stresses on these planes have magnitude of 80 MPa.
Use Mohr's circle to find,
(i) Principal stresses and their planes
(ii) Maximum shears stress and their planes and
(iii) Normal and shear stresses on a plane inclined
Given,
at 450 to 150 MPa stress.
(June2015)
fx = –150 MPa
fy = –100 MPa
q = 80 MPa
If Mohr's circle is drawn to scale, all the quantities can be obtained graphically. However, the
present example has been solved analytically using Mohr's circle.
Construct Mohr's circle with earlier fig
To find Radius of Circle
To find Principal Stress and Principal Planes
fn
–max
= OC + CD
= – 125 – 83.82
= – 208.82 MPa
fn
min
= OC – CE
= – 125 – (–83.82)
Dept of Civil Engg., SJBIT
Page 11
Strength of materials
10CV33
= – 41.18 MPa
From figure
Unit 4
1. Derive a relationship between BM, SF and intensity of load
(June 2014/Jan2015)
RELATIONS BETWEEN LOAD, SHEAR AND MOMENT
Consider a simply supported beam subjected to a Uniformly Distributed Load w/m. Let us
assume that a portion PQRS of length
portion
is cut and taken out. Consider the equilibrium of this
F
M
w
M+M
x
∑V=0
F+F
F)- w x =0
F – (F+
∆
∆
Limit x  0, then =
x
= −
− or F = ∫
Taking moments about section CD for equilibrium
Dept of Civil Engg., SJBIT
Page 12
Strength of materials
10CV33
M-(M+M)+F 
x-(w(x)2 /2) =0
F=
Limit x 0 then F =
Δ
Δ
or M = ∫
Rate of change of Shear Force or slope of SFD at any point on the beam is equal to the intensity
of load at that point.
2. Define Shear Force, Bending moment, SFD and BMD
(June 2013)
Shear Force
The imbalance shear force which shears the beam in a section is called as Shear Force
It is a single vertical force developed internally at any point on the beam to balance the external
vertical forces and keep the point in equilibrium. It is therefore equal to algebraic sum of all
external forces acting to either left or right of the section.
Bending Moment
The imbalance moment, which bends into a circular arc is called as bending moment. It is a
moment developed internally at each point in a beam that balances the external moments due to
forces and keeps the point in equilibrium. It is the algebraic sum of moments to section of all
forces either on left or on right of the section
Shear Forces Diagram (SFD)
The SFD is one which shows the variation of shear force from section to section along the
length of the beam. Thus the ordinate of the diagram at any section gives the Shear Force at that
section.
Bending Moment Diagram (BMD)
The BMD is one which shows the variation of Bending Moment from section to section
along the length of the beam. The ordinate of the diagram at any section gives the Bending
Moment at that section.
Dept of Civil Engg., SJBIT
Page 13
Strength of materials
10CV33
3. Explain the terms : i)Hogging bending moment ii) Sagging bending moment and iii)
Point of contra-flexure
(June2015)
Sagging bending moment
The top fibers are in compression and bottom fibers are in tension.
Hogging bending moment
The top fibers are in tension and bottom fibers are in compression.
Point of Contraflexure
When there is an overhang portion, the beam is subjected to a combination of Sagging and
Hogging moment. The point on the BMD where the nature of bending moment changes from
hogging to sagging or sagging to hogging is known as point of contraflexure. Hence, at point of
contraflexure BM is zero. The point corresponding to point of contraflexure on the beam is
called as point of inflection
Dept of Civil Engg., SJBIT
Page 14
Strength of materials
10CV33
4. Draw SFD, BMD for the beam shown in the fig 3. Also locate the point of contra-flexure.
(Jan 2015)
20kN
3kN/m
2m
2m
RA =39.4kN
4m
2m
RB =58.6kN
Fig 3
39.4
26
19.4
SFD(kN)
5.492m
32.6
132.07
117.6
78.8
BMD(kN-m)
7m
26
Dept of Civil Engg., SJBIT
Page 15
Strength of materials
10CV33
5. Calculate the SF and BM at all the salient points and draw the SFD and BMD for the
beam shown in the fig 4
(June 2013)
30kN/m
100kN
60˚
90 kN-m
A
B
4m
1m
1m
RA =112.73kN
RB =93.87kN
112.73
7.27
3.75m
93.87
93.87
211.8
210.9
93.87
+
BMD(kN-m)
3.87
Dept of Civil Engg., SJBIT
Page 16
Strength of materials
10CV33
Unit 5
1. Define Neutral axis, Section Modulus and Moment of Resistance
(June 2013/Jan2015)
Netural Axis
The layer neither subjected to tension nor to compression. Such a layer is called “Neutral
Layer”. The projection of Neutral Layer over the cross section of the beam is called “Neutral
Axis”.
Section Modulus:Section modulus of a beam is the ratio of moment of inertia of the cross
section of the beam about the neutral axis to the distance of the farthest fiber from neutral
axis.
Moment of Resistance:
The tensile and compressive stresses result in a turning effect about the neutral axis. These
are called moment MT and MC respectively. The chosen beam must be able to resist these
moments with MR (internal moment of resistance) if it is to remain in equilibrium.
2. State the assumptions made in the pure bending






(Dec 2013)
The material is isotropic and homogenous.
The material is perfectly elastic and obeys Hooke's Law i.e., the stresses are within the
limit of proportionality.
Initially the beam is straight and stress free.
Beam is made up of number of layers and they undergo bending independently.
Bending takes place over an arc of a circle and the radius of curvature is very large when
compared to the dimensions of the beam.
Normal plane sections before bending remain normal and plane even after bending.
Dept of Civil Engg., SJBIT
Page 17
Strength of materials
10CV33
3. Derive the equation of theory ofsimple bending w ith usual notation or
Derive a general bending equation =
=
with usual notation (Dec 2013/June2015)
Euler- Bernoulli’s Equation
R
A1
G1
C
D
E1
C
B1
H1
F1
D
C1
D1
Consider two section very close together (AB and CD). After bending the sections will be at A1
B1 and C1 D1 and are no longer parallel. AC will have extended to A1 C1 and B1 D1 will have
compressed to B1D1. The line EF will be located such that it will not change in length. This
surface is called neutral surface and its intersection with Z-Z is called the neutral axis.
The development lines of A'B' and C'D' intersect at a point 0 at an angle of θ radians and the
radius of E1F1 = R.
Let y be the distance(E'G') of any layer H1G1 originally parallel to EF.
Then H1G1/ E 1F1 =(R+y)θ /R θ = (R+y)/R
and the strain
at layer H1G1= = (H1G1'- HG) / HG = (H1G1- HG) / EF
= [(R+y)θ - R θ] /R θ
= y /R.
The relation between stress and strain is σ= E.
Therefore
σ = E. = E. y /R
σ/E=y/ R
Dept of Civil Engg., SJBIT
Page 18
Strength of materials
10CV33
Let us consider an elemental area ‘da ‘at a distance y, from the Neutral Axis.
da
stress developed over elemental area is
y
=
Force developed over elemental area = stress × area =
Moment developed over elemental area about NA= Force × distan ce =
Total Moment developed from all the elements about the NA = ∫
M=
∫
=
=
…………..(2)
=
=
From Eqn (1) and (2), we get
……… …..(A)
4. Compare the flexural strength of the following three beams:
i) I- section 320mm X160mm with 20mm thick flange and 13mm thick web
ii) Rectangle section having depth twice the width
iii) Solid circular section
All the three beam sections have same cross-sectional area
(Jan 2015)
Dept of Civil Engg., SJBIT
Page 19
Strength of materials
10CV33
5. A T section is having a flange of 200mm X 50mm. The web is also 200mm X 50mm. It
is subjected to a bending moment 15kN-m. Draw the bending stress distributing across
the section indicating the salient features
(Dec 2013/June2015)
200mm
50mm
×
=
×
= 162.5
200mm
I=
y1
×
+ (200 × 50 × (225 − 162.5)
50m
×
+ (200 × 50 × (162.5 − 100)
82.29X10 mm4
The Bending moment equation is
=
=
=
=
= 15.94 /
=
= 29.62 /
15.94N/mm2
Yt =87.5mm
Yc = 162.5mm
29.62N/mm2
Dept of Civil Engg., SJBIT
Page 20
Strength of materials
10CV33
Unit 6
1. Define: 1) Slope 2) Deflection, 3) Elastic curve
(June 2013/June2014)
Slope
Angle made by the tangent to the elastic curve with respect to horizontal
Deflection:
Vertical displacement measured from original neutral surface (refer to earlier chapter) to the
neutral surface of the deformed beam.
Elastic Line:
It is a line which represents the deformed shape of the beam. Hence, it is the line along
which the longitudinal axis of the beam bends
2. Derive an expression for the slope and deflection of a simply supported beam carrying a
point load at the centre
(June2013)
Concentrated load P at its Mid-span.
P
A
B
L/2
L
…………………(1)
…………………(2)
At x =0; y =0  C2 =0
At x = L y =0
Dept of Civil Engg., SJBIT
Page 21
Strength of materials
10CV33
Maximum deflection occurs at x = L/2
Substituting the values of x and C1 in equation…. (2)
The negative sign indicates that the deflection is below the undeformed neural axis
3. Establish the equation for slope and deflection for a cantilever beam of length L and carrying a
UDL of w kN /m throughout. Also determine max slope and deflection. EI is constant
(June 2014/Jan2015)
Dept of Civil Engg., SJBIT
Page 22
Strength of materials
10CV33
Boundary conditions relevant to the problem are as follows:
1. At x = L; y = 0
2. At x= L; dy/dx = 0
The second boundary conditions yields
Unit 7
1. List the assumptions made in the theory of torsion
(Dec 2013/June2015)
ASSUMPTIONS IN TORSION THEORY
1. Material is homogenous and isotropic
2. Plane section remain plane before and after twisting i.e., no warpage of planes.
3. Twist along the shaft is uniform.
4. Radii which are straight before twisting remain straight after twisting.
5. Stresses are within the proportional limit.
2. Derive the torsion equation for circular member
=
=
with usual notations
(June 2014/June 2013/Jan2015)
Dept of Civil Engg., SJBIT
Page 23
Strength of materials
10CV33
Consider now the solid circular shaft of radius R subjected to a torque T at one end, the other end
being fixed Under the action of this torque a radial line at the free end of the shaft twists through an
angle q , point A moves to B, and AB subtends an angle ‘ g ' at the fixed end. This is then the angle
of distortion of the shaft i.e the shear strain.
Since angle in radius = arc / Radius
arc AB = Rq
= L g [since L and g also constitute the arc AB]
Thus, g = Rq / L (1)
From the definition of Modulus of rigidity or Modulus of elasticity in shear
Stresses: Let us consider a small strip of radius r and thickness dr which is subjected to shear
stress t'.
The force set up on each element
= stress x area
= τ' x 2p r dr (approximately)
This force will produce a moment or torque about the center axis of the shaft.
Dept of Civil Engg., SJBIT
Page 24
Strength of materials
10CV33
= τ' . 2 p r dr . r
= 2 p τ' . r2. dr
The total torque T on the section, will be the sum of all the contributions.
=
2
′
Since t' is a function of r, because it varies with radius so writing down t' in terms of r from the
equation (1).
Dept of Civil Engg., SJBIT
Page 25
Strength of materials
10CV33
3. A 150 mm diameter solid steel shaft is transmitting 450 kW power at 90rpm. Compute
the maximum shearing stress. Find the change that would occur in the shearing stress, if
the speed were increased to 360rpm
(Dec 2013)
Solution:
=
=
⟹
⟹
= 47746.48
=
−
= 7.2 /
When the speed is increased to 360mm
Tave = 11936.62N-m
= 1.801 /
4. Determine the diameter of a solid circular shaft which will rans
t mit 400 kilowatts at 300
rpm. The maximum shear stress should not exceed 32 N /m
e twist should not be
.th
°
more than 1 in a length of 2 m, assume the modulus of rigi dity sa 90 kN/m2.
(June 2014/June2015)
Solution
⟹
=
=
= 12732.40
−
The diameter due shear stress
=
⟹
12732.4 × 1000
=
32
2
32
d = 126.54 mm
The diameter due to angle of twist
=
⟹
127 2.4 × 1000
32
=
90 × 10 × 0.0174
2000
d = 325 mm
Dept of Civil Engg., SJBIT
Page 26
Strength of materials
10CV33
Unit 8
1. List the assumptions made in Euler’s theory of
long columns
(June 2014/Jan2015)
Solution:
The assumptions made in the analysis are as follows:





The column is long and fails by buckling.
The column is axially loaded.
The column is perfectly straight and the cross sections are uniform (prismatic).
The column is initially free from stress.
The column is perfectly elastic, homogeneous and isotropic.
2. Distinguish short column and long column, Define “slenderness ratio” of a column.
(June 2013/Dec 2013/Jan2015/June2015)
Short Column :
A short column essentially fails by crushing and not by buckling. A column is said to be short, if
le /b 15 or le /rmin 50, where le = effective length, b = least lateral dimension and rmin=
minimum radius of gyration.
Long Column :
A long column essentially fails by buckling and not by crushing. In long columns, the stress at
failure is less than the yield stress. A column is said to be long le/b > 15 or le /rmin> 50.
Slenderness ratio is defined as the ratio of effective length (l e ) of the column to the minimum
radius of gyration (r min ) of the cross section.
Since an axially loaded column tends to buckle about the axis of minimum moment of inertia
(I min), the minimum radius of gyration is used to calculate slenderness ratio.
Further,
, where A is the cross sectional area of column.
Dept of Civil Engg., SJBIT
Page 27
Strength of materials
10CV33
3. Using Euler’s theory, derive an equation for the crippling load of a long column Pinned at
both ends
(June 2013)
Both ends hinged (pinned)
Consider a long column with both ends hinged subjected to critical load P as shown.
P
x
P
Consider a section at a distance x from the origin. Let y be the deflection of the column at this
section. Bending moment in terms of load P and deflection y is given by
We can also write that for beams/columnsthe bending moment is proportional to the curvature
of the beam, which, for sm
all deflection can be expressed as
=
or
=
…………….(2 )
where E is the Young's modulus and I is the moment of Inertia.
Substituting eq.(1) in eq.(2)
Dept of Civil Engg., SJBIT
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Strength of materials
10CV33
This is a second order differential equation, which has a general solution form of
where C1 and C2 are constants. The values of constants can be obtained by applying the
boundary conditions:
(i) y = 0 at x = 0. That is, the deflection of the column must be zero at each end since it is pinned
at each end. Applying these conditions (putting these values into the eq. (3)) gives us the
following results: For y to be zero at x =0, the value of C2 must be zero (since cos (0) = 1).
(i) Substituting y = 0 at x = L in eq. (3) lead to the following.
While for y to be zero at x = L, then either C1 must be zero (which leaves us with no equation at
all, if C1 and C2 are both zero), or
which results in the fact that
Taking least significant value of n, i.e. n = 1
Dept of Civil Engg., SJBIT
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Strength of materials
10CV33
where le =L.
4. A simply supported beam of length 4 m is subjected to a uniformly distributed load of 30
kN/m over the whole span and deflects 15mm at he
t centre. Determine the crippling load
when this beam is used as a column wit h bo th theends hinged
(June 2013)
5
384
=
EI = 6.67 X 109 N-mm2
× .
=
=
×
= 4114.3 9 N
5. A solid round bar 4m long and 50mm in diameter was found to extend by 4.6 mm long
under a tensile load of 50kN. This bar is used a strut with both ends hinged {pinned}.
Determine Euler’s crippling load for the bar andalso safe load taking factor of safety as
4.
(Dec 2014)
Solution
E=
=
=
×
×
=
× ×
×
×
= 221 43.30 /
× .
.
×
.
= 4190.60 N
Psafe = PE/FS = 4190.6/4 = 1047.64 N
6. Find the Euler’s critical load for a hollow cylindrical cast iron column 150 mm external
diameter, 20 mm wall thickness if it is 6 m long with hinges at both ends. Assume
Young’s modulus of cast iron as 80 kN/mm2. Compare this load with given by Rankine’s
formula using Rankine’s constant a = 1/1,600 and fc = 567 N/mm2.
(June 2015)
Dept of Civil Engg., SJBIT
Page 30
Strength of materials
10CV33
Given Data: External diameter = 150 mm, thickness = 20 mm, length = 6 m, E = 80
kN/mm2, a = 1/1,600 and fc = 567 N/mm2.
6
4
2
A0 = 8167.4 1 mm , I = 17.66 X 10 m m
From Euler’s Condition le = l
From Rankine Formulae
=
=
×
=
=
×
×
×
.
.
×
= 387.33
= 406.01
.
Dept of Civil Engg., SJBIT
Page 31